How to determine the radius of convergence of $\sum\frac{1}{n^{\sqrt{n}}}x^n$ without using l'hopital's rule?

I first tried root test, so $|a_n|^{\frac{1}{n}}=\frac{1}{n^{\frac{1}{\sqrt{n}}}}$ which I cannot see what the limit it has.

Then I tried ratio test, so $R=\lim|\frac{a_n}{a_{n+1}}|=\lim|\frac{{(n+1)}^{\sqrt{n+1}}}{n^{\sqrt{n}}}|$ in which I have no idea how to cancel term and determine the limit.

Does anyone have idea?

  • 1
    $\begingroup$ Where would L'Hopital be used here? $\endgroup$
    – user228113
    Jun 4 '17 at 16:54
  • $\begingroup$ What is $x_n$? You probably mean $x^n$. $\endgroup$
    – Mark Viola
    Jun 4 '17 at 17:02


$$n^{1/\sqrt n}=\left(n^{1/2\sqrt n}\right)^2=\left[\left(\sqrt n\right)^{1/\sqrt n}\right]^2\xrightarrow[n\to\infty]{}\ldots$$


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